What is the integral of 16e^{-4x} ?

The integral of 16e^{-4x} is -4e^{-4x}+C

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integral of 16e^{-4x}

\int 16 e^{- 4 x} d x

- 4 e^{- 4 x} + C

Solution steps

\int 16 e^{- 4 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int e^{- 4 x} d x

Apply u-substitution

= 16 \cdot \int - \frac{1}{4} e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 (- \frac{1}{4} \cdot \int e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= 16 (- \frac{1}{4} e^{u})

Substitute back

u = - 4 x

= 16 (- \frac{1}{4} e^{- 4 x})

Simplify

16 (- \frac{1}{4} e^{- 4 x}) : - 4 e^{- 4 x}

= - 4 e^{- 4 x}

Add a constant to the solution

= - 4 e^{- 4 x} + C